tag:blogger.com,1999:blog-6933544261975483399.post5798508359821841707..comments2020-03-28T11:11:19.574-07:00Comments on Go Geometry: Geometry Problem 1314: Equilateral Triangle, Incircle, Inscribed Circle, Tangent, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-71137577218660497562017-02-19T20:46:14.952-08:002017-02-19T20:46:14.952-08:00Let the radius of O = r and that of O1 and O2 = u....Let the radius of O = r and that of O1 and O2 = u.<br /><br />Since &lt; O1AD = 30, r+u = 2(r-u) <br />hence r = 3u......(1)<br /><br />DE^2 = 4{(r+u)^2 - (r-u)^2} = 48u^2 ..(2) from (1)<br /><br />FT^2 = (2r+u)^2 - u^2 = 48u^2 ..(3) also from (1)<br /><br />From (2) and (3) DE = FT<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-60789964891181487072017-02-19T14:31:51.572-08:002017-02-19T14:31:51.572-08:00Join F to A , T to O1, O to M ( midpoint AC)From R...Join F to A , T to O1, O to M ( midpoint AC)From Right tr with angle 30°<br />=&gt; R = 3r =&gt; TF = 4r√3. From AOM =&gt; AC = 6r√3, AD=EC=r√3 =&gt; DE = 4r√3c.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-67306611994744916232017-02-19T13:47:41.678-08:002017-02-19T13:47:41.678-08:00Problem 1314 Suppose the circle with center O in...Problem 1314<br /> Suppose the circle with center O intersects the circle O1 at point L. Bring the common tangent at the point L which intersects AB and AC in P, Q respectively.Draw OK perpendicular in AC (K on AC). The points A,L,O and F are collinear.Is &lt;OAK=30<br />so OA=2OK=2r, and triangle APQ is equilateral, then AO1=2AL/3.But LO=KO, &lt;LOK=60<br />then triangle LKO is equilateral so &lt;LKA=0-60=30=&lt;KAL.Therefore AL=LK=LO=OK=r.<br />So AO1=AO/3 and r1=O1D=OK/3=r/3. In triangle OTF ( &lt;OTF=90) we have <br />FT^2=O1F^2-O1T^2=(2r+r1)^2-r1^2=4r(r+r1)=16r^2/3 (1).Draw O1M//DE (M on OK).<br />Is &lt;O1MO=90 then O1M^2=OO1^2-OM^2=(r+r1)^2-(r-r1)^2=4rr1=4r^2/3. Now <br />DE^2=4DK^2=4O1M^2=16r^2/3 (2).Therefore FT=DE.<br />APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE <br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.com